Chapter 5, Problem 77RE

Solution

a.

To explain: The reason that $\angle A^{\prime }BC$ is a right angle.

The measure of an inscribed angle that intercept an arc of $180°$ is a right angle.

Given information: A triangle $\Delta ABC$ is circumscribed in a circle as shown in the figure below, with diameter $C{A}^{\prime }$ .

  

Figure (1)

Explanation:

According to inscribed angle theorem, the measure of an inscribed angle is the half of the measure of its intercepted arc.

  $\text{measure}\text{of}\angle A^{\prime }BC=\frac{1}{2}\left(\text{measure}\text{of}\text{arc}{A}^{\prime }C\right)$

The measure of intercepted arc inscribed in a circle is $180°$ . So, the measure of arc $A^{\prime }C$ is $180°$ .

  $\begin{array}{c}\text{measure}\text{of}\angle A^{\prime }BC=\frac{1}{2}\left(180°\right)\\ =90°\end{array}$

Thus, $\angle A^{\prime }BC$ is a right angle because the measure of an inscribed angle that intercept an arc of $180°$ is a right angle.

b.

To explain: The reason that $\angle A^{\prime }$ and $\angle A$ are congruent.

The given $\angle A^{\prime }$ and $\angle A$ are congruent because both the angles intercept the same arc $BC$ .

Given information: A triangle $\Delta ABC$ is circumscribed in a circle as shown in the figure below, with diameter $C{A}^{\prime }$ .

  

Figure (1)

Explanation:

From the figure it can be observed that, $\angle A^{\prime }$ intercepts the arc $BC$ and $\angle A$ also intercepts the arc $BC$ .

It is known that if two inscribed angles intercept the same arc then the angle are said to be congruent. So, $\angle A^{\prime }$ and $\angle A$ are congruent

Thus, $\angle A^{\prime }$ and $\angle A$ are congruent because both the angles intercept the same arc $BC$ .

c.

To explain: The reason that $\sin A^{\prime }=\frac{a}{d}$ if $a$, $b$ and $c$ are the sides opposite angles $A$, $B$ and $C$ in $\Delta ABC$ .

The value $\sin A^{\prime }=\frac{a}{d}$ because $\Delta A^{\prime }BC$ is a right triangle.

Given information: A triangle $\Delta ABC$ is circumscribed in a circle as shown in the figure below, with diameter $C{A}^{\prime }$ . In $\Delta ABC$, $a$, $b$ and $c$ are the sides opposite angles $A$, $B$ and $C$ . The diameter of the circle is $d$ .

  

Figure (1)

Explanation:

In part (a), it is proved that $\angle A^{\prime }BC$ is a right angle. So, $\Delta A^{\prime }BC$ is a right triangle.

Diameter $A^{\prime }C=d$ and side $BC=a$ opposite to $\angle A$ .

Apply the trigonometric ratio of sine in $\Delta A^{\prime }BC$ .

  $\begin{array}{c}\sin A^{\prime }=\frac{BC}{A^{\prime }C}\\ =\frac{a}{d}\end{array}$

Thus, $\sin A^{\prime }=\frac{a}{d}$ because $\Delta A^{\prime }BC$ is a right triangle.

d.

To explain: The reason that $\frac{\sin A}{a}=\frac{1}{d}$ if $a$, $b$ and $c$ are the sides opposite angles $A$, $B$ and $C$ in $\Delta ABC$ .

The value $\frac{\sin A}{a}=\frac{1}{d}$ because $\angle A^{\prime }$ and $\angle A$ are congruent.

Given information: A triangle $\Delta ABC$ is circumscribed in a circle as shown in the figure below, with diameter $C{A}^{\prime }$ . In $\Delta ABC$, $a$, $b$ and $c$ are the sides opposite angles $A$, $B$ and $C$ . The diameter of the circle is $d$ .

  

Figure (1)

Explanation:

In part (c), it is proved that $\sin A^{\prime }=\frac{a}{d}$ .

In part (b), $\angle A^{\prime }$ and $\angle A$ are congruent.

Substitute $A$ for $A^{\prime }$ in $\sin A^{\prime }=\frac{a}{d}$ .

  $\begin{array}{l}\sin A=\frac{a}{d}\\ \frac{\sin A}{a}=\frac{1}{d}\end{array}$

Thus, value $\frac{\sin A}{a}=\frac{1}{d}$ because $\angle A^{\prime }$ and $\angle A$ are congruent.

e.

To check: The ratio $\frac{\sin B}{b}$ and $\frac{\sin C}{c}$ also equal to $\frac{1}{d}$, and explain the reason.

Yes, $\frac{\sin B}{b}$ and $\frac{\sin C}{c}$ both are equal to $\frac{1}{d}$ by applying of the law of Sines.

Given information: A triangle $\Delta ABC$ is circumscribed in a circle as shown in the figure below, with diameter $C{A}^{\prime }$ . In $\Delta ABC$, $a$, $b$ and $c$ are the sides opposite angles $A$, $B$ and $C$ . The diameter of the circle is $d$ .

  

Figure (1)

Explanation:

In $\Delta ABC$, apply the law of Sines.

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

From part (d), it is proved that $\frac{\sin A}{a}=\frac{1}{d}$ . So,

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$

Thus, $\frac{\sin B}{b}$ and $\frac{\sin C}{c}$ both are equal to $\frac{1}{d}$ by applying of the law of Sines.